modern geometry fall 2011

MODERN GEOMETRYFALL 2011

P O R T F O L I OB Y : A L E X A N D R I A C R O O MP R O F E S S O R : D R . D A V I D J A M E S

DESCRIPT

ION OF

PROJECT

As a senior at Howard University, I enrolled in a directed reading course with Dr. David James. This project is a way to document my progress in the course. This document is a work in progress that will continued to be completed throughout the semester. I would like to especially thank Dr. James for being willing to teach this course. His enthusiasm has brought much joy to the learning process.

HOW WAS MODERN

GEOMETRY DEVELO

PED?

Check out this video!

TABLE OF CONTENTSThe Geometries

EuclideanHyperbolicSpherical

AxiomsDistanceIncidenceLinesReal Ray

Theorems & ProofsUnique Middle TheoremTriangle Inequality

THE GEO

METRIES

E X A M P L E S OF F

I VE S

Y S T E M S

EUCLIDEAN GEOMETRY

The Euclidean plane points and lines includes: Coordinates - set of points is an ordered pair Equation of lines

Nonvertical y= mx+ b Vertical x= a

Distance Formula e(AB)=

HYPERBOLIC GEOMETRYConsists of all points inside (but not on) the unit circle in the

Euclidean plane. That is all (x, y)Lines in H are the chords of the circleDistance in the Hyperbolic Plane is calculated:

Where M, N are endpoints of chords, and A, B are points on a line

SPHERICAL GEOMETRYS(r): notation for spherical plane surface of a sphere of radius rS- set of all (x, y, z) with Definition: Great circles- intersection of sphere with plane that cuts it in

halfHow is distance defined in S?The distance between two points is the length of the minor arc of the

great circle (line) through A and B

AXIOMS

A D E S C R I P T I ON O

F ON E O

R MO

R E MA T H S T R U C T U R E S C O

N S I S T I N G

O F T E R MS , D

E F I N I T I ON S , A S S U M

P T I ON S , T H E O

R E MS , A N D

P R OP O

S I T I ON S

AXIOMS OF DISTANCEFor all points P and Q

1. Positivity: PQ 2. Definiteness: iff 3. Symmetry:

AXIOMS OF INCIDENCEIn a system1. There are at least two different lines2. Each line contains at least two different points3. Each two different points lie in at least one line4. Each two different points P, Q, with lie in at most one line

THREE AXIOMS FOR THE LINEDefinition: Point B lies between points A and C (written A-B-C) provided

that A, B, C are different collinear points, and AB+BC = ACUnique Middle Theorem follows (see theorems and proofs)

Betweenness of Points AxiomIf A, B, and C are different collinear points and if then there exists a

betweenness relation among A, B, and C.

Triangle inequality theorem follows (see theorems & proofs)

THREE AXIOMS FOR THE LINEQuadrichotomy Axiom for Points

If A, B, C, X are distinct, collinear points, and if A-B-C, then at least one of following must hold:

X-A-B, A-X-B, B-X-C, OR B-C-X

THREE AXIOMS FOR THE LINENontriviality AxiomFor any point A on a line there exists a point n with

REAL RAY AXIOMDefinition: Let A, B, be points with The ray is the set

For any ray and any real number with there is a point X in with

Leads to Theorem 8.2 ( see theorems & proofs)

THEOREMS & PR

OOFS

K E Y T H E O R E MS A N D P O S T U L A T E S F O

R WO

R K I N G I N MO

D E R N

G E OM

E T R Y

UNIQUE MIDDLE THEOREMIf A-B-C, then both B-A-C and A-C-B are false.Proof:By definition of betweenness, A, B, C are distinct collinear point and AB + BC = AC.Assume B-A-C. Then,

BA + AC = BCBA + AB + BC = BC2AB + BC= BC (by symmetry axiom AB =BA)2AB=0

Thus, AB= 0. This implies that 0 + BC = AC By positive definiteness this would make A=B, contradicting the assumption that they are distinct points

Therefore, B-A-C is false

A-C-B is likewise false.

TRIANGLE INEQUALITY

If A, B, C are distinct and collinear then AB + BC ≥ AC

Proof of Triangle Inequality

Case 1: 𝐴𝐵+ 𝐵𝐶≥ 𝜔

Then by Betweenness of Points Axiom, there is some betweenness relationship between A, B, C.

Without loss of Generality consider two case: 1) A-B-C and 2) A-C-B

1) Since A-B-C by definition AB+BC = AC

Therefore it holds that 𝐴𝐵+ 𝐵𝐶≥ 𝐴𝐶

2) For A-C-B,

𝐴𝐶+ 𝐵𝐶= 𝐴𝐵

𝐴𝐶= 𝐴𝐵− 𝐵𝐶 ≤ 𝐴𝐵+ 𝐵𝐶

Therefore it holds that 𝐴𝐵+ 𝐵𝐶 ≥ 𝐴𝐶

Case 2: 𝐴𝐵+ 𝐵𝐶 > 𝜔

For E, G, H, and M, 𝜔= ∞. Do as longs as 𝐴𝐵+ 𝐵𝐶= 𝑥 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑥∈𝐑 there will always be a 𝜔 greater.

Thus ∃ a betweenness relation fro A, B, C such that by case 1 it holds that 𝐴𝐵+ 𝐵𝐶 ≥ 𝐴𝐶

For A,B, C ∈ H, if 𝐴𝐵+ 𝐵𝐶> 𝜔 then 𝐴𝐵+ 𝐵𝐶 > 𝜋𝑟 = 𝜔

AB + BC must also be the length of the major axis.

It follows AC must be the length of the major axis.

Therefore 𝐴𝐶≤ 𝜋𝑟 = 𝜔< 𝐴𝐵+ 𝐵𝐶

Therefore it holds that 𝐴𝐶≤ 𝐴𝐵+ 𝐵𝐶. Thus, the proposition is true.

THIS IS STILL A WORK IN PROGRESS!

Special Thanks to :Dr. David James

AndThe Howard University Mathematics Department